0
1449/1 – 1 hour 15 minutes
40 objective questions
1449/2 – subjective
Section A - 11 compulsory
            questions
Sect...
3 marks
Example
    ξ                    B

             A                   C


                 II     III IV   V   VI


       ...
ξ                                 B

               A                              C


                        II         ...
3.     Identify the shaded region
                     U
     ( II, III, IV       III, IV, V, VI )   U    IV, V
          ...
SETS
          Shade the Region
  (a) P ∩ R '          (b)   P ∩ Q ∪ R'
                                     Q
           ...
P ∩ R'




P ∩ Q ∪ R'
Linear inequalities
Shade the region that satisfies the inequalities
     Know how to sketch a straight line
     Applicat...
Shade the region that satisfies the three
inequalities y ≤ 2 x + 8 , y ≥ x and y < 8

                                  y=...
Know how to sketch the straight line y =8
From , y = 2 x + 8 , y-intercept = 8
                      y = 2x + 8
          ...
y = 2x + 8
                   y=x
8




          less marks
4 marks
Solid Geometry & Volume
Combination of two
solids
1. Determine the two solids involved
2.Choose the operations + or -
3.Wr...
Try this
A hemisphere PQR has taken out from the
cylinder.
Find the volume of the remaining solid.

 P             R      ...
The diagram shows a solid cone with radius 9 cm
and height 14 cm. A cylinder with radius 3 cm
and height 7 cm is taken out...
Diagram 3 shows a solid cone with radius 9 cm and
height 14 cm. A cylinder with radius 3 cm and height 7
cm is taken out o...
1
Vcone = πr 2 h
        3 Stress on correct values when substituting
   1  22 
          ( 9 ) ( 14 )
                ...
Lines and Planes
      in 3D

    4 marks
Lines and Planes in 3D

      IMPORTANT NOTES :

      3. SKETC therig a le
                H    ht- ng d
         tria le...
LINES AND PLANES IN 3 DIMENSION
Name the angle between the line CE and the plane EFGH
Arrange the line and plane in two ro...
Name the angle between the plane DGK and the base DEFG
        WON
DK = GKW O N
                                Slash the ...
V

WON
WON
Non-slash alphabet
Slash alphabet
                                                      6

                    ...
A                  B

                             Name the angle between the
WON
E
                 F
                   ...
Try this..
        H         U
                                 G


  E                      F




                  T
   ...
Find the angle between the plane JFE
and the plane DEF.



      L                                J

                     ...
L                           J

                                5

   F                            D
        5
            ...
5
tan JMD   =
                13 − 5
                  2      2




∠   JMD       = 22.62    o

                          ...
5M A R K S
MATHEMATICAL REASONING
Is the following sentence a statement ?
Give your reason.
     9 + 2 = 2 −9
 Statement             ...
Make a conclusion for the number sequence
below

  2 −1= 0
   0
                          F u ll m a r k s
  2 −1=1
   1
 ...
n
2 ─ 1 , n = 0 , 1 , 2 , 3 , ..
 n
2 ─ 1 , n =0,1,2,3,
 n
2 ─ 1 , n =0,1,2,
 n
2 ─ 1




     Ie s s marks
Write two implications from this
          compound statement

     3   p = 5 if and only if p = 125

If         3   p = 5...
Complete the following argument
Premise1 : If 4x = 16 , then x = 4

Premise 2 : x ≠ 4

Conclusion : 4 x ≠ 16
SIMULTANEOUS LINEAR
     EQUATIONS
   Elimination method

   Substitution method

     Matrix method

     4 M AR KS
Solve the simultaneous linear equations
               1
                 p − 2q = 9
               3
              5 p + ...
Elimination method
      1                               p − 6q
        p − 2q = 9 same deno min ator        =9
      3   ...
Substitution method
    1                               p − 6q
      p − 2q = 9 same deno min ator        =9
    3        ...
Matrix method        1      p  9 
                        − 2
                               = 
              ...
THE STRAIGHT LINE – 6 MARKS
REMEMBER :
               y1 − y2
• Gradient m =
               x1 − x2
• Equation of a line
 ...
THE STRAIGHT LINE

REMEMB ER :

3. Parallel lines , same gradient   m1 = m2
4. Perpendicular lines , the product of their
...
THE STRAIGHT LINE


REMEMB ER :

5. x-intercept , substitute   y=0
6. y-intercept , substitute   x=0
Important notes

x-intercept   = − 12

x = − 12

x-intercept is   ( − 12 , 0 )
In Diagram 2, O is the origin, point R lies on the x-
axis and point P lies on the y-axis. Straight line PU
is parallel to...
(a) PU is parallel to the x-axis.
      Find the value of its y-intercept from              x + 2y = 14.
                 ...
Find the equation of the straight line ST and
hence, state its x-intercept.
               y                              ...
hence, state its x-intercept.

             1
      y = − x−4
             2
      x − int ercept ∴ y = 0
                ...
QUADRATIC EQUATIONS

      4 MARKS
QUADRATIC EQUATION

REARRANGE TO GENERAL FORM OF
QUADRATIC EQUATION

   ax + bx + c = 0
      2


  Factorise   (     )(  ...
Solve the quadratic equation
              3n + 6n = 7 ( 1 − 2n )
                 2


3n + 6n = 7 (1 − 2n ) ⇒ 3n + 6n − 7...
2k − 5
                                 2
Solve the quadratic equation          = 3k
                                 3

 ...
MATRICES
NOTES
1. When the matrix has no inverse
       ad − bc = 0
2. MATRIX FORM

3. Formula of the inverse matrix

4. S...
 5 − 9
The inverse of matrix   
                         2 − 3  is
                                
                ...
Use the inverse formula
                     1           − 3 9
inverse =                             
                ...
Calculate the value of x and the value of y by
using matrix method

           2 x − 3 y = −14
           − x + 3 y = 13
F...
Write the inverse formula IN   FRONT
                           1  d − b
                                      
      ...
 2 − 3  x   − 14 
               = 
        −1 3   y   13  
                        

         x ...
Wrong arrangement

        x   2 − 3     − 14 
        
        y   − 1 3  =  13 
                        ...
CIRCLES : Perimeter and Area

1. Use the correct formulae

3. Substitute with the correct values.

                22
    ...
P

        S          9 cm

     7 cm   W                 U
                                         6 cm
       T
       ...
P

       S          9 cm

    7 cm   W                 U
                                        6 cm
       T
          ...
(a).
   Perimeter = PU + UV + VQ + ( PQ – ST ) + SWT

      30       22                      180       22  7  
 = 9+...
(b).
        Area


         1            180 22  7  2  30 22 2  
       =  × 12 × 9  −      × ×  −       × ...
1            180 22  7  2  30 22 2  
a) =  × 12 × 9  −     × ×  −       × × 6 
     2           360 7 ...
Diagram 4 shows two sectors ORST and OUV with the same
centre O. RWO is a semicircle with diameter RO and RO=2OV.
ROV and ...
a ) perimeter = RO + OV + RST + TU + UV
                            120       22        
              =14 + 7 +       ...
b) area = OUV + ORST − OWR
            60 22 2   120 22                 2
         =       × ×7  +              × ×...
4 marks
PROBABILITY II


         n( A )
 P( A) =
         n( S )
A group of 5 boys and 4 girls take part in a study on the
  type of plants found in a reserved forest area. Each
  day, tw...
To choose a boy, the probability,

               5                             4
  P (1 boy ) =
      st
                ...
Two boys then exempted from writing the report on the
second day

     Both boys-2 boys                         Both girls...
Society                Number of Student
                               boy             girl
      Science                ...
5 marks
GRADIENT AND AREA UNDER A
                  GRAPH

                                     Distance
Speed                Cons...
GRADIENT AND AREA UNDER A GRAPH

 REMEMBER :
 2. Le ng th o f time is to tal time take n

 2. AREA o f trape zium
 3. Dis ...
Speed (ms-1)



          21                           (t ,21)




          9


           1                            (...
State the length of time, in s, that the particle moves
   with uniform speed.

                                    12 − 5...
-1
(b) Calculate the rate of change of speed, in ms , in the
   first 5 seconds.
    Speed (ms-1)


   21                 ...
Calculate the value of t, if the total distance
travelled for t seconds is 148 metres.
  Speed (ms-1)


 21               ...
SUGGESTED TIME


 15 – 20 MINUTES
  PER QUESTION
12 (a) Complete table 1 in the answer space for the
    equation y =2x2-x-3.                   ( 2 marks)

x          -2  ...
12 (a) Complete table 1 in the answer space for the
equation y =2x2 – x - 3.                      ( 2 marks)




x    -2  ...
12
     GRAPHS OF FUNCTIONS
     1. Fill in the blanks in the table
           y = 2 x − 12 x + 3
                       3...
2. Draw the graphs of functions

Scales and the range of x are given

    −3 ≤ x ≤ 5
    Plot the points accurately




  ...
through




the curve must be smooth, passing
through each point
losing marks




Using your own scale
 12 marks – 1 mark
-10
-15
-20
e12c. e a n s w e r in t h e a n s w e r s p a c e s p r
   th



                                   x =1.5
12c.      The first equation from (a)
              y = x − 6x + 5
                        2

         The second equation...
12c.     The first equation from (a)
             y = x − 6x + 5
                      2

        The second equation from...
12d.
TRANSFORMATIONS III

A Combined Transformation RS means
transformation S followed by transformation R.


- Use the right t...
y      E                  H

                        4
                                   F
                              ...
Enlargement

                          centre( 2,6 )
6
     E        H           Scale factor 3
4
     J                  ...
losing marks




       Please use the right term
rotation
enlargement              correct direction
correct centre
k : sf : ratio ( m: n)
                         Reflect...
STATISTICS spm

2006 FREQUENCY
POLYGON
2005 HISTOGRAM
2004 HISTOGRAM
2003 OGIVE
STATISTICS


Mean , median , modal class
2.Frequency Polygon
3.Histogram
4.Ogive
5.Information of the graph
Marks     Midpoint   Frequency
20 – 24      22          5
25 - 29      27          7
30 - 34      32          8
35 - 39   ...
MEAN TABLE
 x             fx
92       92 x 4 = 368            10257
97       97 x 10 = 970
                          mean ...
These are students’ answers

Mean =
        0 x 38 + 4 x 43 + 6 x 48 +
12 x 53 + 9 x 58 + 5 x 63 + 6 x 68 + 8 x 73

      ...
Frequency polygon table
                        frequency   midpoint
              height        f         x
             ...
FREQUENCY POLYGON
                        POLIGON KEKERAPAN

F           30
R
                                            ...
POLIGON KEKERAPAN
                  FREQUENCY POLYGON
      F
            30
      R
      E     25                       ...
TABLE FOR HISTOGRAM
          frequency midpoint   Upper boundary
CLASS        f           x
36 – 40      0          38   ...
F u ll m a r k
40.5   45.5   50.5   55.5   60.5   65.5   70.5   75.5


                                      F u ll m a r k s
41 - 45


      46 - 50


      51 - 55


      56 - 60


      61 - 65


    61 - 70


      71 - 75
F u ll m a r k
Le s s m a r


o u ld h a v e g a p
41 - 45 46 - 50 51 - 55       61 - 65        71 - 75
                      56 - 60        61 - 70


                      ...
Table for OGIVE

           frequency   Cumulative   Upper boundary
CLASS         f        frequency
36 – 40       0      ...
Cumulative                             F u ll m a r k
       120
frequency          OGIF
             100
 LONGGOKAN
 KEKE...
Cumulative
frequency
       120         OGIF
             100
 LONGGOKAN
 KEKERAPAN




              80

              60...
Cumulative                    OGIF
frequency
                                                           N o m a rk
       ...
Third Quartile
                           3
                           4
                           1
Median( Second Quart...
Ogive Of Time Taken For 100 Students To Complete
                                                     Their Compositions

...
INTERQUARTILE
                                     Ogive Of Time Taken For 100 Students To Complete             RANGE
    ...
Information of the graph


     50 students took 60 minutes to complete
     their composition.

      Interquartile range...
PLANS AND ELEVATIONS

CORRECT SHAPE

Satisfy the given CONDITIONS

MEASUREMENT MUST BE
ACCURATE
 LATERAL INVERSION is not ...
15b(ii).




           H id d e n lin e
losing marks
15




  Case 1:
 Double line
  Bold line
15a.


       Case 2 :
       Sizes – Bigger or Smaller
Case 3:
extension



            Case 4 :
             gap
15




     Case 5:
Not a right angles
Draw a full scale
 i. The plan of the solid
ii. the elevation of the solid as viewed from Y
iii. the elevation of the soli...
F u ll m a r k
N o m a rk
EARTH
AS A SPHERE
Longitude
                     N

 MG 0°

              40°
               20°




     40° E
                     S
     ...
TWO meridians form a GREAT
          circle
            N
  30 w               150 E




            S
U


135 W       45 E




        S
P(60 N, 30 W ) and Q are two points on the surface of the earth
where PQ is the diameter of the parallel latitude of P and...
J( 30 S, 80 E ) and K are two points on the earth where
JK is the diameter of the earth. The location of K is

A. ( 30 S, ...
a)P is a point on the surface of the earth such
   that JP is the diameter of the earth. State the
   position of P.
b) Ca...
16. Interpretation of question – sketch the
    earth

    o
  40 W                                    o
      J          ...
16b.      JM = MK
        ∴ angle at the centre JOM = 90 o
        Greenwich
 o
40 W                                      ...
16c.
                                      JK = 80 × 60
                                         = 4800
       J          ...
16b.       JMK route
        ∴ angle at the centre JOM = 180   o


                        = 180 × 60 × cos 50   o

   o
 ...
Reminders
   Calculate , find , solve ,
    – all steps are clearly shown

   State – only the answer is required

   U...
Reminders
Basic Mathematical Skills such as
 addition, division, subtraction,
 multiplication.
Algebraic and Trigonometr...
Reminders
 Allsteps must be clearly shown.
 Read the instructions and questions very
  carefully .
 The answer must be ...
Mathematics Keynotes 2
Mathematics Keynotes 2
Mathematics Keynotes 2
Mathematics Keynotes 2
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Transcript of "Mathematics Keynotes 2"

  1. 1. 1449/1 – 1 hour 15 minutes 40 objective questions 1449/2 – subjective Section A - 11 compulsory questions Section B - 4 out of 5 questions 2 Hours 30 minutes SCIENTIFIC CALCULATOR GEOMETRIC AL SET
  2. 2. 3 marks
  3. 3. Example ξ B A C II III IV V VI I Shade the set ( A ∩ B ) ∪ C 1. Label each part with a Roman number
  4. 4. ξ B A C II III IV V VI I 2. Identify the shaded region U (A B) U C U ( II, III, IV III, IV, V, VI ) U IV, V
  5. 5. 3. Identify the shaded region U ( II, III, IV III, IV, V, VI ) U IV, V III, IV U IV, V III, IV, V 4. Shade the region mark with III, IV, V U (A B) U C
  6. 6. SETS Shade the Region (a) P ∩ R ' (b) P ∩ Q ∪ R' Q Q P P R Intersection R Union Com plim of ent
  7. 7. P ∩ R' P ∩ Q ∪ R'
  8. 8. Linear inequalities Shade the region that satisfies the inequalities Know how to sketch a straight line Application of y-intercept Understand the inequality sign < , > for dashed line and ≤ , ≥ for solid line
  9. 9. Shade the region that satisfies the three inequalities y ≤ 2 x + 8 , y ≥ x and y < 8 y=x 0 y = 2x + 8
  10. 10. Know how to sketch the straight line y =8 From , y = 2 x + 8 , y-intercept = 8 y = 2x + 8 y=x 8 y<8 Full marks
  11. 11. y = 2x + 8 y=x 8 less marks
  12. 12. 4 marks
  13. 13. Solid Geometry & Volume Combination of two solids 1. Determine the two solids involved 2.Choose the operations + or - 3.Write the correct formulae 4.Substitute the values of r, h, d 22 Use π= 7
  14. 14. Try this A hemisphere PQR has taken out from the cylinder. Find the volume of the remaining solid. P R Cylinder - hemisphere 2 3 πr h − πr 2 3 Q 8 cm 22 2 22 ×5× 5×8 − × 5× 5× 5 7 3 7 2 366 10 cm 3
  15. 15. The diagram shows a solid cone with radius 9 cm and height 14 cm. A cylinder with radius 3 cm and height 7 cm is taken out of the solid. Calculate the volume, in cm3 , of the remaining solid. 22 Use π = 7
  16. 16. Diagram 3 shows a solid cone with radius 9 cm and height 14 cm. A cylinder with radius 3 cm and height 7 cm is taken out of the solid. 14 cm Write the formula first 7 cm 1 2 Vcone = πr 3 cm 3 9 cm Vcylinder = π r h 2
  17. 17. 1 Vcone = πr 2 h 3 Stress on correct values when substituting 1  22  ( 9 ) ( 14 ) 2 =  3 7  Remaining solid =1188 = 1188 – 198 Vcylinder =πr h 2 = 990  22  Stress on correct values ( 3 ) ( 7 ) 2 = when substituting  7  =198
  18. 18. Lines and Planes in 3D 4 marks
  19. 19. Lines and Planes in 3D IMPORTANT NOTES : 3. SKETC therig a le H ht- ng d tria le ng 2 Id ntify thea lea na eit . e ng nd m n g le b e t w e e n a lin e a n d a p la A n g le b e t w e e n t w o p la n e s
  20. 20. LINES AND PLANES IN 3 DIMENSION Name the angle between the line CE and the plane EFGH Arrange the line and plane in two rows Find out the same alphabet Look at the line and the plane in the diagram C Write C in the first box B Look at c, choose Which One is the Nearest to C A D (slashed alphabet) 6 C E Look at C, WON E F G Choose W O N H (Non-slashed alphabets) F G Look at the diagram θ 6 C E C E G 6 H Tan θ= 6 Draw 3 boxes K2 √72 θ E 36 + 36 G θ= 35.26°
  21. 21. Name the angle between the plane DGK and the base DEFG WON DK = GKW O N Slash the same alphabet Write K in the first box So, choose the midpoint of DG H Look at K, choose W O N Slashed- alphabet K D G K 8 G D E F G N θ 6 Look at the diagram D F Look at the diagram Look at the diagram 12 M K N M E Look at K, choose W O N K Tan θ= 6 EK = FK Non-slashed alphabets choose M 12 θ= 26.57° θ N M
  22. 22. V WON WON Non-slash alphabet Slash alphabet 6 D C θ 8 A 10 B Calculate the angle between the plane AVC and the plane BVC A V C A Tan θ = 10 B V C 8 A C B θ= 51.34° θ C B
  23. 23. A B Name the angle between the WON E F plane ACGE with the plane WON DCGH A C G E D θ C D C G H H G A C D between A & E, choose either one and write the alphabet in the first box ( for rectangle only)
  24. 24. Try this.. H U G E F T P N L M R Name the angle between the plane LUM with the plane LPNM Answer : URT
  25. 25. Find the angle between the plane JFE and the plane DEF. L J 5 F D 5 M 13 5 E
  26. 26. L J 5 F D 5 M 13 5 E Identify the angle ∠ JMD
  27. 27. 5 tan JMD = 13 − 5 2 2 ∠ JMD = 22.62 o o ' atau 22 37
  28. 28. 5M A R K S
  29. 29. MATHEMATICAL REASONING Is the following sentence a statement ? Give your reason. 9 + 2 = 2 −9 Statement yes Not accepted answer A statement. It is a false statement.
  30. 30. Make a conclusion for the number sequence below 2 −1= 0 0 F u ll m a r k s 2 −1=1 1 3 d o ts 2 −1= 3 2 2 −1= 7 3 2 − 1, n = 0 ,1, 2 , 3 , ... n
  31. 31. n 2 ─ 1 , n = 0 , 1 , 2 , 3 , .. n 2 ─ 1 , n =0,1,2,3, n 2 ─ 1 , n =0,1,2, n 2 ─ 1 Ie s s marks
  32. 32. Write two implications from this compound statement 3 p = 5 if and only if p = 125 If 3 p = 5 , then p = 125 F u ll m a r k If p = 125 , then 3 p =5 3 p = 5 , then p = 125 n o m a rk If p = 125 , 3 p =5
  33. 33. Complete the following argument Premise1 : If 4x = 16 , then x = 4 Premise 2 : x ≠ 4 Conclusion : 4 x ≠ 16
  34. 34. SIMULTANEOUS LINEAR EQUATIONS Elimination method Substitution method Matrix method 4 M AR KS
  35. 35. Solve the simultaneous linear equations 1 p − 2q = 9 3 5 p + 6q = −9 When there is a fraction, you must have the same denominator first p − 6q =9 3 p − 6q = 27
  36. 36. Elimination method 1 p − 6q p − 2q = 9 same deno min ator =9 3 3 ⇒ p −6q = 27 1 5 p + 6q = −9 2 1 +2 6 p = 18 r r e c t o p e r a t io n ∴ p=3 ⇒ 3 − 6q = 27 ∴ q = −4
  37. 37. Substitution method 1 p − 6q p − 2q = 9 same deno min ator =9 3 3 ⇒ p −6q = 27 ∴ p = 27 +6q C o r r e c t o p e r a t io n 1 5 p + 6q = −9 2 substitute 1 into 2 5 ( 27 + 6q ) + 6q = − 9 ⇒ p = 27 + 6 (− 4 ) 135 + 30q + 6q = −9 ∴ p=3 ∴ q = −4
  38. 38. Matrix method 1  p  9   − 2  =   3  q   − 9  5 6     C o r r e c t m a t r ix f o r m  p 1  6 2 9   =    1   q  1 − 5 − 9    × 6 − 5 × ( − 2)  3 3  p  3   =   ∴ p = 3 , q = −4  q   −4    
  39. 39. THE STRAIGHT LINE – 6 MARKS REMEMBER : y1 − y2 • Gradient m = x1 − x2 • Equation of a line y = mx + c x y + = 1 a b
  40. 40. THE STRAIGHT LINE REMEMB ER : 3. Parallel lines , same gradient m1 = m2 4. Perpendicular lines , the product of their gradients = − 1 m1m2 = −1
  41. 41. THE STRAIGHT LINE REMEMB ER : 5. x-intercept , substitute y=0 6. y-intercept , substitute x=0
  42. 42. Important notes x-intercept = − 12 x = − 12 x-intercept is ( − 12 , 0 )
  43. 43. In Diagram 2, O is the origin, point R lies on the x- axis and point P lies on the y-axis. Straight line PU is parallel to the x-axis and straight line PR is parallel to the straight line ST. The equation of straight line PR is x + 2y = 14. y (a) Find the value of its • • U P y-intercept from x + 2y = 14. x + 2y = 14 S • (b) Find the equation of the O • R x straight line ST and hence, state its x-intercept. • T (2,-5)
  44. 44. (a) PU is parallel to the x-axis. Find the value of its y-intercept from x + 2y = 14. y P • • U 2 y = − x + 14 x 14 x + 2y = 14 y=− + S 2 2 • • x O R x y = − +7 2 • T (2,-5) y-intercept! y-intercept =7
  45. 45. Find the equation of the straight line ST and hence, state its x-intercept. y From part (a), we have • • U x P 2 y = − x + 14 → y = − + 7 2 1 x + 2y = 14 Therefore, gradient ST , m = − S 2 • 1 • x Substitute m = − and po int (2, −5) in O R 2 y = mx + c,  1 −5 =  −  (2) + c • T (2,-5)  2 c = −5 + 1 = −4 Thus, equation ST is 1 y = − x−4 2
  46. 46. hence, state its x-intercept. 1 y = − x−4 2 x − int ercept ∴ y = 0 1 4=− x 2 x = −8 x-intercept = – 8
  47. 47. QUADRATIC EQUATIONS 4 MARKS
  48. 48. QUADRATIC EQUATION REARRANGE TO GENERAL FORM OF QUADRATIC EQUATION ax + bx + c = 0 2 Factorise ( )( )=0 State the values of x
  49. 49. Solve the quadratic equation 3n + 6n = 7 ( 1 − 2n ) 2 3n + 6n = 7 (1 − 2n ) ⇒ 3n + 6n − 7 + 14n = 0 2 2 ∴ 3n + 20n − 7 = 0 2 Fa c to rs mus t be g iv e n b y u s in g w h o le ( 3n − 1 ) ( n + 7 ) = 0 n u m b e rs 1 ∴ n= , −7 3
  50. 50. 2k − 5 2 Solve the quadratic equation = 3k 3 2k − 5 = 9k 2 2k − 9k − 5 = 0 2 ( 2k + 1) ( k − 5) = 0 1 k =− , k =5 2
  51. 51. MATRICES NOTES 1. When the matrix has no inverse ad − bc = 0 2. MATRIX FORM 3. Formula of the inverse matrix 4. State the value of x and of y
  52. 52.  5 − 9 The inverse of matrix   2 − 3  is    1  − 3 9   q 5  p  Find the values of p and q
  53. 53. Use the inverse formula 1  − 3 9 inverse =    − 2 5 5 × (−3) − (−9) × 2   1  − 3 9 Compare with the =   − 2 5  given inverse matrix 3  1  − 3 9 ∴ p= 3   q 5  p  q= −2
  54. 54. Calculate the value of x and the value of y by using matrix method 2 x − 3 y = −14 − x + 3 y = 13 Form a matrix equation  2 − 3  x   − 14    − 1 3   y  =  13          
  55. 55. Write the inverse formula IN FRONT 1  d − b   − c a  F u ll m a r k s ad − bc    x  1  3 3   − 14   =    y  3  1 2   13         x   − 1 ∴ x = −1  =    y  4      y=4
  56. 56.  2 − 3  x   − 14     =   −1 3   y   13         x  1  3 3   − 14   =   y  3  1 2   13          x   − 1  =    y  4      le s s m a r k s
  57. 57. Wrong arrangement  x   2 − 3  − 14     y   − 1 3  =  13           x  1  2 3  − 14   =    y  3  1 3  13         x  1  3 − 3   − 14   =   y  3  − 1 2   13    n o m a rk      x  − 14  1  3 3   =   y  13  3  1 2          
  58. 58. CIRCLES : Perimeter and Area 1. Use the correct formulae 3. Substitute with the correct values. 22 π = 7
  59. 59. P S 9 cm 7 cm W U 6 cm T 30O R Q 6 cm V 6 cm a) Find the perimeter of the shaded region b) Calculate the area of the shaded region
  60. 60. P S 9 cm 7 cm W U 6 cm T 30O R Q 6 cm V 6 cm Perimeter = PU + UV + VQ + ( PQ – ST ) + SWT Area = area of triangle – area of hemisphere – area of the sector
  61. 61. (a). Perimeter = PU + UV + VQ + ( PQ – ST ) + SWT  30 22   180 22  7   = 9+ × 2 × × 6  + 6 + ( 9 − 7) +   360 × 2 × 7 ×  2     360 7     218 = 7 1 31 7 31.14
  62. 62. (b). Area 1    180 22  7  2  30 22 2   =  × 12 × 9  − × ×  − × × 6  2   360 7  2   360 7   709 = 28 9 25 28 25.32
  63. 63. 1    180 22  7  2  30 22 2   a) =  × 12 × 9  −  × ×  − × × 6  2   360 7  2   360 7  709 = A n s w e r s a r e in t h e w ro n g 28 a ns w e r s pa c e s b)  30 22   180 22  7   = 9+ × 2 × × 6  + 6 + ( 9 − 7) +   360 × 2 × 7 ×  2     360 7     218 = 7
  64. 64. Diagram 4 shows two sectors ORST and OUV with the same centre O. RWO is a semicircle with diameter RO and RO=2OV. ROV and OUT are straight lines. OV=7cm and angle UOV= 60˚ Calculate (a) perimeter of the whole diagram , (b) area of the shaded region. S T 7 c U W m o 6 0V R O 14 c 7 c m m
  65. 65. a ) perimeter = RO + OV + RST + TU + UV  120 22  =14 + 7 +  × 2 × ×14  + 7  360 7 Stress on the correct values when substituting   60 22  + × 2× × 7  360 7  2 = 64 3
  66. 66. b) area = OUV + ORST − OWR  60 22 2   120 22 2 = × ×7  +  × × 14   360 7   360 7  Stress on the correct values when substituting  180 22 2  − × ×7   360 7  = 154
  67. 67. 4 marks
  68. 68. PROBABILITY II n( A ) P( A) = n( S )
  69. 69. A group of 5 boys and 4 girls take part in a study on the type of plants found in a reserved forest area. Each day, two pupils are chosen at random to write a report. • Calculate the probability that both pupils chosen to write the report on the first day are boys. (b) Two boys has written the report on the first day. They are then exempted from writing the report on the second day. Calculate the probability that both pupils chosen to write the report on the second day are the same gender.
  70. 70. To choose a boy, the probability, 5 4 P (1 boy ) = st P (2 boy ) = nd 9 8 5 4 5 P (both boys ) = × = 9 8 18
  71. 71. Two boys then exempted from writing the report on the second day Both boys-2 boys Both girls-2 girls n(all girls ) n(all boys ) P (one girl ) = P (one boy ) = n(all pupils ) n(all pupils ) 4 3 P (1 girl ) = st P (1 boy ) = st 7 7 3 2 P (2nd girl ) = P (2nd boy ) = 6 6 4 3 2 3 2 1 Thus, P(2boys ) = × = Thus, P(2boys ) = × = 7 6 7 7 6 7 1 2 3 P(both pupils) = + = 7 7 7
  72. 72. Society Number of Student boy girl Science 3 5 4 Consumer 6 7 a. If two students were chosen at random from the science society, calculate the probability that both are girls GxG 5 x 4 = 5 8 7 14 b. If two students were chosen at random from the group of boys, calculate the probability that both boys came from the same society S x S or C x C 3/9 x2/8 + 6/9 x 5/8 =1/2
  73. 73. 5 marks
  74. 74. GRADIENT AND AREA UNDER A GRAPH Distance Speed Constant/ Uniform speed Object stops Time Time m = rate of change of speed m = Rate of change of distance = speed/time = distance/ time = acceleration / deceleration = speed Area under the graph is the distance
  75. 75. GRADIENT AND AREA UNDER A GRAPH REMEMBER : 2. Le ng th o f time is to tal time take n 2. AREA o f trape zium 3. Dis tanc e -time g raph , g radie nt = s pe e d e quivale nt to the rate o f c hang e o f dis tanc e 4. S pe e d -time g raph , g radie nt = ac c e le ratio n e quivale nt to the rate o f c hang e o f s pe e d
  76. 76. Speed (ms-1) 21 (t ,21) 9 1 (t ,0) O 5 12 t Time (s) (a) State the length of time, in s, that the particle moves with uniform speed. (b) Calculate the rate of change of speed, in ms-1 , in the first 5 seconds.
  77. 77. State the length of time, in s, that the particle moves with uniform speed. 12 − 5 = 7 Speed (ms-1) 21 (t ,21) Time (s) 9 1 (t ,0) O 5 12 t
  78. 78. -1 (b) Calculate the rate of change of speed, in ms , in the first 5 seconds. Speed (ms-1) 21 (t ,21) 9 1 (t ,0) O 5 12 t Time (s) 9 −1 8 gradient of the green straight line = = 12 − 5 7
  79. 79. Calculate the value of t, if the total distance travelled for t seconds is 148 metres. Speed (ms-1) 21 (t ,21) 9 1 (t ,0) O 5 12 t Total distance= area under the graph Stress on the correct values when substituting 1 1 148 = (1 + 9 ) 5 + ( 7 × 9 ) + ( 9 + 21) ( t −12 ) 2 2 10t = 180 t = 18
  80. 80. SUGGESTED TIME 15 – 20 MINUTES PER QUESTION
  81. 81. 12 (a) Complete table 1 in the answer space for the equation y =2x2-x-3. ( 2 marks) x -2 -1 -0.5 1 2 3 4 4.5 5 y 7 -2 -2 3 12 33 42 12 (a) In the table 1 , find the value of m and the value of n for the equation y = 2 x 3 − 12 x + 3 ( 2 marks) x -3 -2 -1 0 1 2 y -15 m 13 3 -7 n
  82. 82. 12 (a) Complete table 1 in the answer space for the equation y =2x2 – x - 3. ( 2 marks) x -2 -1 -0.5 1 2 3 4 4.5 5 y 7 -2 -2 3 12 33 42 0 25
  83. 83. 12 GRAPHS OF FUNCTIONS 1. Fill in the blanks in the table y = 2 x − 12 x + 3 3 -2 2 x m n y 11 −5
  84. 84. 2. Draw the graphs of functions Scales and the range of x are given −3 ≤ x ≤ 5 Plot the points accurately Can use flexible curve
  85. 85. through the curve must be smooth, passing through each point
  86. 86. losing marks Using your own scale 12 marks – 1 mark
  87. 87. -10 -15 -20
  88. 88. e12c. e a n s w e r in t h e a n s w e r s p a c e s p r th x =1.5
  89. 89. 12c. The first equation from (a) y = x − 6x + 5 2 The second equation from (c) x = 7x −4 2 eliminate variables that have indices , 2 1 3 x ,x , x
  90. 90. 12c. The first equation from (a) y = x − 6x + 5 2 The second equation from (c) x = 7x −4 2 eliminate variables that have indices y − x = x − 6 x + 5 − (7 x − 4) 2 2 y = x − x − 6x − 7x + 5 + 4 2 2 y = −13 x + 9
  91. 91. 12d.
  92. 92. TRANSFORMATIONS III A Combined Transformation RS means transformation S followed by transformation R. - Use the right terminologies - Start the answer with the right transformation - No short form - Describe in full the transformation – with the correct properties.
  93. 93. y E H 4 F y=3 D G 2 Rotation C 180o A B o centre ( 0,3 ) -4 -2 2 4 Describe in full the transformation PQ
  94. 94. Enlargement centre( 2,6 ) 6 E H Scale factor 3 4 J M 2 G K 2 4 6 8 -2 L
  95. 95. losing marks Please use the right term
  96. 96. rotation enlargement correct direction correct centre k : sf : ratio ( m: n) Reflection at a point A enlargement correct scale factor reflection enlargement rotation correct centre enlargement rotation angle translation
  97. 97. STATISTICS spm 2006 FREQUENCY POLYGON 2005 HISTOGRAM 2004 HISTOGRAM 2003 OGIVE
  98. 98. STATISTICS Mean , median , modal class 2.Frequency Polygon 3.Histogram 4.Ogive 5.Information of the graph
  99. 99. Marks Midpoint Frequency 20 – 24 22 5 25 - 29 27 7 30 - 34 32 8 35 - 39 37 10 40 - 44 42 6 45 - 49 47 4 50 - 54 52 2
  100. 100. MEAN TABLE x fx 92 92 x 4 = 368 10257 97 97 x 10 = 970 mean = 96 102 102 x 26 = 2652 107 107 x 24 = 2568 112 112 x 17 = 1904 = 106.84 117 117 x 9 = 1053 122 122 x 4 = 488 127 127 x 2 = 254 10257
  101. 101. These are students’ answers Mean = 0 x 38 + 4 x 43 + 6 x 48 + 12 x 53 + 9 x 58 + 5 x 63 + 6 x 68 + 8 x 73 0 + 4 + 6 + 12 + 9 + 5 + 6 + 8 = 58.5
  102. 102. Frequency polygon table frequency midpoint height f x 90 - 94 4 92 95 - 99 10 97 100 - 104 26 102 105 - 109 24 107 110 - 114 17 112 115 - 119 9 117 120 - 124 4 122 125 - 129 2 127 JUMLAH 96
  103. 103. FREQUENCY POLYGON POLIGON KEKERAPAN F 30 R F u ll m a r k s 25 KEKERAPAN E 20 Q U 15 E 10 N C 5 Y 0 87 92 97 102 107 112 117 122 127 132 MIDPOINT TITIK TENGAH
  104. 104. POLIGON KEKERAPAN FREQUENCY POLYGON F 30 R E 25 le s s m a r k s KEKERAPAN Q 20 U 15 E N 10 C 5 Y 0 7 7 2 2 2 2 7 87 92 97 12 13 10 10 11 11 12 MIDPOINT MARKAH
  105. 105. TABLE FOR HISTOGRAM frequency midpoint Upper boundary CLASS f x 36 – 40 0 38 40.5 41 – 45 4 43 45.5 46 – 50 6 48 50.5 51 – 55 12 53 55.5 56 – 60 10 58 60.5 61 – 65 5 63 65.5 66 – 70 5 68 70.5 71 – 75 8 73 75.5 JUMLAH 50
  106. 106. F u ll m a r k
  107. 107. 40.5 45.5 50.5 55.5 60.5 65.5 70.5 75.5 F u ll m a r k s
  108. 108. 41 - 45 46 - 50 51 - 55 56 - 60 61 - 65 61 - 70 71 - 75 F u ll m a r k
  109. 109. Le s s m a r o u ld h a v e g a p
  110. 110. 41 - 45 46 - 50 51 - 55 61 - 65 71 - 75 56 - 60 61 - 70 no mark
  111. 111. Table for OGIVE frequency Cumulative Upper boundary CLASS f frequency 36 – 40 0 0 40.5 41 – 45 4 4 45.5 46 – 50 6 10 50.5 51 – 55 12 22 55.5 56 – 60 10 32 60.5 61 – 65 5 37 65.5 66 – 70 5 42 70.5 71 – 75 8 50 75.5 ∑ 50
  112. 112. Cumulative F u ll m a r k 120 frequency OGIF 100 LONGGOKAN KEKERAPAN 80 60 40 20 0 5 5 5 5 5 5 .5 .5 .5 .5 4. 9. 4. 9. 4. 9. 84 89 94 99 10 10 11 11 12 12 SEMPADAN ATAS Upper boundary
  113. 113. Cumulative frequency 120 OGIF 100 LONGGOKAN KEKERAPAN 80 60 le s s m a r k 40 20 0 5 5 5 5 5 5 .5 .5 .5 .5 4. 9. 4. 9. 4. 9. 84 89 94 99 10 11 11 12 10 12 SEMPADAN ATAS Upper boundary
  114. 114. Cumulative OGIF frequency N o m a rk 110 100 LONGGOKAN KEKERAPAN 90 80 70 60 50 40 30 20 10 0 0 0 0 0 5 5 5 80 90 85 95 10 10 11 11 12 12 13 limit MARKAH
  115. 115. Third Quartile 3 4 1 Median( Second Quartile) 2 1 First Quartile 4
  116. 116. Ogive Of Time Taken For 100 Students To Complete Their Compositions 105 100 3 95 100× 90 4 85 80 Finding the third quartile Cumulative Frequency 75 70 65 100 60 55 Finding the median 2 50 (second quartile) 45 40 35 100 30 Finding the first quartile 25 4 20 15 10 5 0 0 20 40 60 80 100 Time ( minutes )
  117. 117. INTERQUARTILE Ogive Of Time Taken For 100 Students To Complete RANGE Their Compositions 105 100 Q3 − Q1 3 Q3 95 100× 90 4 85 80 Finding the third quartile Cumulative Frequency 75 70 65 100 60 Finding the median 2 55 50 (second quartile) 45 40 35 Q1 100 30 Finding the first quartile 4 25 20 15 10 5 0 0 20 40 60 80 100 Time ( minutes )
  118. 118. Information of the graph 50 students took 60 minutes to complete their composition. Interquartile range is 20 minutes Median is 60 minutes n f o r m a t io n m u s t c o m e f r o m t h e g
  119. 119. PLANS AND ELEVATIONS CORRECT SHAPE Satisfy the given CONDITIONS MEASUREMENT MUST BE ACCURATE LATERAL INVERSION is not accepted (SONGSANG SISI TIDAK DITERIMA)
  120. 120. 15b(ii). H id d e n lin e
  121. 121. losing marks
  122. 122. 15 Case 1: Double line Bold line
  123. 123. 15a. Case 2 : Sizes – Bigger or Smaller
  124. 124. Case 3: extension Case 4 : gap
  125. 125. 15 Case 5: Not a right angles
  126. 126. Draw a full scale i. The plan of the solid ii. the elevation of the solid as viewed from Y iii. the elevation of the solid as viewed from X X Y
  127. 127. F u ll m a r k
  128. 128. N o m a rk
  129. 129. EARTH AS A SPHERE
  130. 130. Longitude N MG 0° 40° 20° 40° E S 60° E
  131. 131. TWO meridians form a GREAT circle N 30 w 150 E S
  132. 132. U 135 W 45 E S
  133. 133. P(60 N, 30 W ) and Q are two points on the surface of the earth where PQ is the diameter of the parallel latitude of P and Q. The position of point Q is A. ( 60 N, 150 W) C. (60 S, 150 E ) B. ( 60 N 150 E) D. (60 S, 150 W ) N P Q 150 E 30 W 60 60 S
  134. 134. J( 30 S, 80 E ) and K are two points on the earth where JK is the diameter of the earth. The location of K is A. ( 30 S, 100E) C. ( 30 N, 100W ) B. ( 30 S, 80 E) D. ( 30 N, 80 W) N 80° E 100° W K 300 300 J S
  135. 135. a)P is a point on the surface of the earth such that JP is the diameter of the earth. State the position of P. b) Calculate the value of x, if the distance from J to K measured along the meridian is 4200 nautical miles. c)Calculate the value of y, if the distance from J due west to L measured along the common parallel of latitude and then due south to M. f) If the average speed for the whole flight is 600 knots, calculate the time taken for the whole flight.
  136. 136. 16. Interpretation of question – sketch the earth o 40 W o J K 140 E o 50 O
  137. 137. 16b. JM = MK ∴ angle at the centre JOM = 90 o Greenwich o 40 W o J o K 140 E 40 o 50 M o o 50 E , 50 N
  138. 138. 16c. JK = 80 × 60 = 4800 J K o 80 o o 50 50 O
  139. 139. 16b. JMK route ∴ angle at the centre JOM = 180 o = 180 × 60 × cos 50 o o 40 B o J 180 K Greenwich M 6942.106 ∴ average speed = 13 = 534
  140. 140. Reminders  Calculate , find , solve , – all steps are clearly shown  State – only the answer is required  Unit / label – must be correct if written
  141. 141. Reminders Basic Mathematical Skills such as addition, division, subtraction, multiplication. Algebraic and Trigonometric skills Formulae and its applications, Formulae and its substitutions. Round off only at the last answer line.
  142. 142. Reminders  Allsteps must be clearly shown.  Read the instructions and questions very carefully .  The answer must be in the lowest form, to 4 significant figures and to 2 decimal places.  Master the calculator  Do not sleep during the exam!
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